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MEASUREMENT OF e/m OF THE ELECTRON
Purpose
1.
To study the effect of electric and magnetic fields on charged particles.
2.
To measure the charge to mass ratio of an electron.
Theory
When an electron moves in a magnetic field B whose direction is perpendicular to the velocity v
of the electron, it is acted on by a Lorentz force F perpendicular to B and v with a magnitude
given by:
F = e vB
(1)
wheree is magnitude of the charge of the electron. This force causes the particle to move in a
circle in the plane perpendicular to the field.
Applying Newton’s 2nd law for the circular motion of electron gives
e vB = m
v2
r
(2)
where r is the radius of the circle and m is the mass of the electron.
If the electron has been accelerated from rest through a potential difference V, then the kinetic
energy acquired is equal to the loss of potential energy.
1
m v 2 = eV
2
(3)
Combining equations (2) and (3) gives
e
2V
= 2 2
m B r
(4)
Thus, when the accelerating voltage V (in volts), the magnetic field B (in Tesla), and the radius
of the circular path r(in meter) are known, the value of the e/m can be computed and is given in
units of Coulomb/kg by equation (4).
Apparatus
The Cathode ray tube used in this experiment is a special one in which the path of a beam of
electrons can be observed directly. The tube is filled with helium gas at a low pressure. Electrons
emitted by the heated cathode are accelerated by the potential difference applied between the
cathode and anode cylinder. Some of the electrons come out in a narrow beam through a circular
hole in the center of the cylinder. This emission is then focused into a narrow beam by the grid of
the tube. When electrons of sufficient high kinetic energy leaving the cathode collide with
heliumatoms a fraction of the atoms will be ionized. Upon combination of these ions with stray
electrons, the helium-arc spectrum is emitted and the characteristic green color of the
undispersed visible light is observed.
Figure 1
A homogenous magnetic field is produced in the region of the cathode ray tube by a current
through two circular coils (see Figure 1). Whenever a charged particle, such as an electron in the
present experiment, is emitted, the force F given by equation (1) acts on the particle and will
deflect it. The particle, therefore, moves under the influence of this force which has a constant
magnitude but whose direction is always at right angle to the velocity of the particle. The orbit of
the particle is, therefore, a circle.
Calculation of the magnetic field
The magnetic field produced at the position of the electron beam by a current I flowing through
the coils must be calculated. For the equipment used in this experiment, there are two circular
coils with N turns on each, the coils being connected to contribute equally to the field at the
center. This arrangement is known as a pair of Helmholtz Coils and has the advantage that the
field is uniform in the region near the center. For this particular geometry and constructional
details of the apparatus, the relationship between B and Iis given by:
B=
0.716 µ0 N I
R
(5)
whereµo is the permeability constant which is equal to 4π × 10−7T.m/A , N = 130 turns and R is
the radius of the coils = 15 cm.
Procedure
1.
Use a magnetic compass to determine the North-South direction at the location of the
laboratory. Then set the equipment so that the plane of the Helmholtz Coils lies in the
plane of the earth's magnetic field. Why is this step important? Explain.
2.
The circuit has already been connected for you. The power supplies to the filament, coil,
and anode are contained in one pack. Note the following, however:
Figure 2
(i)
The black knob on the left controls the anode voltage, and thedisplay above this
knob reads the accelerating voltage V (0−500 volts).
(ii)
The black knob on the rightcontrols the current through the Helmholtz coils,and
the display above this knob reads the current I.
(iii)
The diameter of the beam 2ris measured by allowing the beam to fall on one of
the phosphor coated cross bars, which allows the direct measurement of the
diameter, as shown in Figure 3.
(iv)
The CR Tube power supply uses high voltage (500 V) and is capable of
delivering high current (350 mA), which can be dangerous.
Electron Gun
= 2 cm
2rr=5
6
37cm
Electron Paths
Phosphor
coated
crossbars
4cm
8 cm 5 cm
Figure 3
3.
Switch on the power supply. This will automatically turn on the filament current. Allow
one or two minutes for the filament to heat up.
4.
Turn the anode voltage knob to apply about 150 volts to the anode. This is the
accelerating voltage V. A fine greenish-colored beam will now be observed.
5.
Switch on the current to the coils. Adjust the current and observe how the beam bends
over and eventually circles back on itself. It may be necessary to rotate the whole tube to
achieve this condition.
6.
Set the current at some value about 1.8 A. Adjust the anode voltage to give an
accelerating voltage V between 150 and 250 volts so that the beam falls on the phosphorcoated crossbars. Before taking your measurements, however, make sure that with the
current value chosen you will be able to bend the beam to fall on the 5 cm cross bar. If
not, then increase the coil current slightly and decrease the anode voltage until the beam
is observed to fall on the cross bar. Keep this value of current I constant in the rest of the
experiment. Record the diameter of the orbit (5 cm) and the accelerating voltage V in
Table 1. Assume 5% error in the ammeter and voltmeter readings.Record the
current as well.
∆I=
I=
7.
Adjust the anode voltage to get the circles of diameters6, 7, 8, 9, 10 and 11 cm and record
the corresponding values of V in Table 1. Complete Column 3 as well.
8.
After finishing your measurements in step 7, turn the current through the coils to zero,
reverse the current leads and increase the current. Observe that the beam is deflected in
the opposite direction. Why does the beam deflect in the opposite direction?
Table 1
2r (cm)
∆V
V
r2
5.0
6.0
7.0
8.0
9.0
10.0
11.0
Data analysis
1.
Use equation (5) to calculate the magnetic field B.
B=
Write the expression for
Therefore,∆B =
∆B
=
B
∆(r2)
e/m
∆(e/m)
2.
Estimate the uncertainty in r to be the thickness of the crossbar.
∆(r) =
Write the expression for
∆(r 2 )
=
r2
Therefore,∆(r2) =
Complete Columns 4 and 5 in Table 1.
3.
Use equation (4) to calculate the ratio e/m.
Write the expression for
∆ (e / m )
=
e/m
Therefore,∆(e/m) =
Complete Columns 6 and 7 in Table 1.
4.
Calculate the average value of e/m.
e/m =
5.
±
The accepted value of e/m is 1.7×1011 C/kg. Calculate the percent difference between the
experimental and accepted values of e/m.
Is the percent difference smaller or larger than your percent error in measuring e/m?
6.
What are the sources of error in this experiment?